Find the root of the perpendicular from point 2- 3- 2 to the line 4 - x2 - y6 - 1 - z3 also find perpendicular distance from the point to the line-

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Points in Different Quadrants

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Question

Find the root of the perpendicular from point $(2, 3, 2)$ to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$ also find perpendicular distance from the point to the line.

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Solution

$\begin{matrix} \frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} \frac{x - 4}{- 2} = \frac{y}{6} = \frac{z - 1}{- 3} F r o m t h e f i g u r e A \equiv (2, 3, 2) B \equiv (- 2 r + 4, 6 r, - 3 r + 1) D i r e c t i o n r a t i o o f A B = (- 2 r + 2, 6 r - 3, - 3 r - 1) dot product of drs of AB and line = 0 - 2 (- 2 r + 2) + 6 (6 r - 3) - 3 (- 3 r - 1) = 0 49 r = 19 r = \frac{19}{49} F o o t o f p e r p e n d i c u l a r = (\frac{- 38}{49} + 4, \frac{114}{49}, \frac{- 57}{49} + 1) = (\frac{158}{49}, \frac{114}{49}, \frac{- 12}{49}) P e r p e n d i c u l a r d i s tan c e = \sqrt{{(\frac{158}{49} - 2)}^{2} + {(\frac{114}{49} - 3)}^{2} + {(\frac{- 12}{49} - 2)}^{2}} = \frac{\sqrt{3600 + 1089 + 121}}{49} = \frac{\sqrt{16, 789}}{49} \end{matrix}$

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Find the root of the perpendicular from point (2,3,2) to the line 4−x2=y6=1−z3 also find perpendicular distance from the point to the line.

Find the root of the perpendicular from point $(2, 3, 2)$ to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$ also find perpendicular distance from the point to the line.